Maximum XOR Queries - Google Top Interview Questions
Problem Statement :
You are given a list of non-negative integers nums and a two-dimensional list of integers queries where each element contains [x, limit].
Return a list such that for each query [x, limit], we find an e in nums such that e ≤ limit and XOR(e, x) is maximized. If there's no such e, use -1.
Constraints
n ≤ 100,000 where n is the length of nums
m ≤ 100,000 where m is the length of queries
Example 1
Input
nums = [2, 4, 8]
queries = [
[3, 5],
[2, 0]
]
Output
[4, -1]
Explanation
For the first query, we can use 2 or 4 in nums. 2 ^ 3 = 1 while 4 ^ 3 = 7 so we pick 4 which yields the
bigger XOR. In the second query, there's no number that's less than or equal to 0, so we set it to -1.
Solution :
Solution in C++ :
struct TrieNode {
int lo = INT_MAX;
TrieNode* children[2]{};
};
class Solution {
public:
vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
TrieNode* root = new TrieNode();
for (int num : nums) {
TrieNode* p = root;
for (int i = 30; i >= 0; --i) {
int nxt = (num & (1 << i)) ? 1 : 0;
if (!p->children[nxt]) p->children[nxt] = new TrieNode();
p = p->children[nxt];
p->lo = min(p->lo, num);
}
}
vector<int> ret;
for (auto q : queries) {
int x = q[0], limit = q[1];
int ans = 0;
TrieNode* p = root;
for (int i = 30; i >= 0; --i) {
if (x & (1 << i)) {
if (p->children[0]) {
p = p->children[0];
} else if (!p->children[1] || (p->children[1]->lo > limit)) {
ret.emplace_back(-1);
break;
} else {
p = p->children[1];
ans ^= (1 << i);
}
} else {
if (p->children[1] && (p->children[1]->lo <= limit)) {
p = p->children[1];
ans ^= (1 << i);
} else if (!p->children[0]) {
ret.emplace_back(-1);
break;
} else {
p = p->children[0];
}
}
if (i == 0) ret.emplace_back(ans);
}
}
return ret;
}
};
vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
return (new Solution())->solve(nums, queries);
}
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] nums, int[][] qs) {
int n = nums.length;
List<Map<Integer, Integer>> ls = new ArrayList<>();
for (int i = 0; i < 32; i++) {
Map<Integer, Integer> mp = new HashMap<>();
for (int x : nums) {
// set.add(x >> i);
mp.put(x >> i, Math.min(x, mp.getOrDefault(x >> i, Integer.MAX_VALUE)));
}
ls.add(mp);
}
Arrays.sort(nums);
int[] res = new int[qs.length];
Arrays.fill(res, -1);
if (n == 0)
return res;
int cnt = 0;
for (int[] q : qs) {
int v = q[0];
int limit = q[1];
int ans = 0;
boolean find = false;
for (int i = 31; i >= 0; i--) {
Set<Integer> ts = ls.get(i).keySet();
Map<Integer, Integer> mp = ls.get(i);
// if(i == 0)System.out.println(ts);
if ((v | (1 << i)) == v) { // this is one
int tk = ans >> i;
if (!ts.contains(tk)) {
ans |= 1 << i;
}
// System.out.println(tk+" "+i+" "+ts+" "+ans);
} else { // we want to get index to be 1
int now = ans | (1 << i);
int tk = now >> i;
/// if we need this index but no that index
if (ts.contains(tk) && mp.get(tk) <= limit) {
ans = now;
}
// System.out.println(tk+" "+i+" "+ts+" "+ans);
}
}
if (ans <= limit && ans >= nums[0]) {
res[cnt] = ans;
}
cnt++;
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums, queries):
nums.sort()
def query(x, limit):
start = 0
stop = bisect_right(nums, limit)
num = 0
for i in range(31, -1, -1):
i_th_bit = 1 << i
plus = num + i_th_bit
if nums[start] >= plus:
num = plus
elif nums[stop - 1] >= plus:
cut = bisect_left(nums, plus, start, stop)
if x & i_th_bit:
stop = cut
else:
start = cut
num = plus
return num
return [query(x, limit) if nums and nums[0] <= limit else -1 for x, limit in queries]
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